Background for Schrödinger's equation

Taking all of the occupancies to be $f=1$, lecture defined the unconstrained objective function for finding multiple solutions to Schrödinger's equation as

$$E=-\frac{1}{2} {\mbox{Tr\,}}\left(W^\dagger L W U^{-1}\right) + \vec V^\dagger {\bf J}^\dagger {\bf O}{\bf J}\vec n,$$

where

$$U \equiv W^\dagger {\bf O}W,$$

$$\vec n \equiv {\mbox{diag\,}}\left( {\bf I}W U^{-1} W^\dagger {\bf I}^\dagger \right),$$

and $\vec V$ is the vector of sample values of the potential on the real space grid.

A minor rearrangement of these expressions, more convenient for our purposes, is

$$E = -\frac{1}{2} {\mbox{Tr\,}}\left(W^\dagger L W U^{-1}\right) + \tilde V^\dagger \vec n$$ (1)

$$U \equiv W^\dagger {\bf O}W$$

$$\vec n \equiv {\mbox{diag\,}}\left( \left({\bf I}W U^{-1}\right) ({\bf I}W)^\dagger \right)$$

$$\tilde V \equiv {\bf J}^\dagger {\bf O}{\bf J}\vec V,$$

where we have regrouped quantities under $\dagger$'s, used the fact that ${\bf O}$ is Hermitian ( ${\bf O}^\dagger={\bf O}$), and defined a ``dual'' set of potential coefficients $\tilde V$ which can be directly combined with $\vec n$ to form the potential energy.

Because $W$ is a complex matrix rather than a real vector, the most convenient form for expressing the gradient of $E$ with respect to $W$ is to form a matrix of the same dimensions as $W$, with each element set equal to the partial derivative of $E$ with respect to the complex conjugate of the corresponding element of $W$,

$$\left[ \nabla_W E \right]_{\alpha,n} \equiv \frac{\partial E}{\partial W_{\alpha,n}^*}.$$ (2)

As derived in lecture, the gradient of $E$ then becomes

$$\nabla_W E = \left( H W - {\bf O}W U^{-1} W^\dagger H W \right) U^{-1},$$ (3)

where $U$ is defined as above and

$$H \equiv -\frac{1}{2} L + {\bf I}^\dagger \left({\mbox{Diag\,}}\tilde V\right) {\bf I}.$$ (4)

Note that with gradients expressed as matrices as in Eq. (3), the following (ultimately!) simple expression may be used to compute the directional derivative of the energy $E$ along the direction $dW$,

$$dE \equiv \sum_{\alpha,n} \left[{\mbox{Re\,}}(dW_{\alpha,n}) \, \frac{\part... ...{\alpha,n}) \, \frac{\partial E}{\partial {\mbox{Im\,}}(dW_{\alpha,n})} \right]$$

$$= \sum_{\alpha,n} {\mbox{Re\,}}\left[ \left( {\mbox{Re\,}}(dW_{\alp... ...})} + i \frac{\partial E}{\partial {\mbox{Im\,}}(W_{\alpha,n})} \right) \right]$$

$$= {\mbox{Re\,}}\sum_{\alpha,n} dW_{\alpha,n}^* 2 \left(\frac{\partial E}{\partial W_{\alpha,n}^*}\right)$$

$$= 2\, {\mbox{Re\,}}{\mbox{Tr\,}}dW^\dagger \nabla_W E.$$ (5)

In the last week of class (time permitting), we will explain the following final note. This isn't needed for computing energies but only if you are interested in the actual eigenstates. In the above, H is a matrix representation of the ``Hamiltonian,'' so that the product $H W$ corresponds to the action of the left-hand side of the eigenvalue equation, ``LHS(W)''. The transformation which turns the unnormalized $W$ into the normalized $Y$ and then the final eigensolutions $\Psi$ is

$$Y \equiv W U^{-1/2}$$ (6)

$$\Psi \equiv Y D,$$ (7)

where $D$ diagonalizes the matrix

$$\mu \equiv Y^\dagger H Y,$$ (8)

according to

$$D^\dagger \mu D = {\mbox{diag\,}}\vec \epsilon,$$ (9)

where $\vec \epsilon$ are the final eigenvalues.